Consider a simply supported beam AB, with span ‘L’ and subjected to an eccentric point load (W), as shown in figure.

To draw SFD and BMD we need RA and RB. By taking moment of all the forces about point ‘A’, we get;

RB . L – W. a = 0

⇒ RB = W.a/L

Also, from condition of static equilibrium Σ Fy = 0

i.e., RA + RB – W = 0

RA = W – Wa/L = W(L – a) / L

⇒ RA = W.b / L

Consider a section (X – X’) at a distance x from end ‘A’.

Shear force = Total unbalanced vertical force on either side of the section

Fx = + RA = + W.b / L

The Fx remains constant between A and C.

(The sign is taken to be positive because the resultant force is in upward direction on the left hand side of the section).

By taking a section between C and B, we get

Fx = + RA – W = + W.b / L – W = W/L (b – L) = – Wa / L

Now, the bending moment between A and C,

Bending moment = Algebraic sum of moments caused by vertical loads acting on either side of the section

⇒ Mx = + RA . x

(The sign is taken to be positive because the load creates sagging).

The bending moment between C and B can be obtained by taking section between C and B at a distance x from A.

Mx = + RA.x – W (x – a) = + Wb/L. x – W(x – a)

To draw SFD and BMD, x is carried from 0 to L.

Since, shear force is not dependent on x the SFD is a rectangle with constant ordinate Wb/L but if changes both sign and magnitude at point C. In between A and C, shear force is positive and between C and B it is negative.

The bending moment is a function of x and its values can be obtained from equations (4) and (6).

MA = + Wb/L.0 = 0

MC = + Wb/L.a = + Wab/L

MB = + Wb.L/L – W(L – a) = Wb – WL + Wa

= W(b + a) – WL

= WL – WL

= 0

The BMD is therefore a triangle with a maximum ordinate of + Wab / L, under the point loading at ‘C’.

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